Measurable Function Zero A.E. iff Absolute Value has Zero Integral/Corollary
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Theorem
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f: X \to \overline \R$ be a non-negative integrable function.
Let $A, B \in \Sigma$ have $A \subseteq B$.
Then:
- $\ds \int_A f \rd \mu = \int_B f \rd \mu$
- $f \times \chi_{B \setminus A} = 0$ $\mu$-almost everywhere.
Proof
We can write:
- $B = A \cup \paren {B \setminus A}$
From Integral of Positive Measurable Function over Disjoint Union, we have:
- $\ds \int_B f \rd \mu = \int_A f \rd \mu + \int_{B \setminus A} f \rd \mu$
Since:
- $\ds \int_B f \rd \mu = \int_A f \rd \mu$
we get:
- $\ds \int_{B \setminus A} f \rd \mu = 0$
From the definition of the $\mu$-integral over $A$, we have:
- $\ds \int \paren {f \times \chi_{B \setminus A} } \rd \mu$
From Measurable Function Zero A.E. iff Absolute Value has Zero Integral, we have:
- $\ds \int \paren {f \times \chi_{B \setminus A} } \rd \mu = 0$ if and only if $f \times \chi_{B \setminus A} = 0$ $\mu$-almost everywhere.
So:
- $\ds \int_A f \rd \mu = \int_B f \rd \mu$
- $f \times \chi_{B \setminus A} = 0$ $\mu$-almost everywhere.
$\blacksquare$